Respuesta :
Given:
Volume of the unknown monoprotic acid (HA) = 25 ml
To determine:
The concentration of the acid HA
Explanation:
The titration reaction can be represented as-
HA + NaOH → Na⁺A⁻ + H₂O
As per stoichiometry: 1 mole of HA reacts with 1 mole of NaOH
At equivalence point-
moles of HA = moles of NaOH
For a known concentration and volume of added NaOH we have:
moles of NaOH = M(NaOH) * V(NaOH)
Thus, the concentration of the unknown 25 ml (0.025 L) of HA would be-
Molarity of HA = moles of HA/Vol of HA
Molarity of HA = M(NaOH)*V(NaOH)/0.025 L
Answer:
The concentration of monoprotic acid on titration against NaOH was found to be 0.36 M.
Explanation:
The reaction of monoprotic acid (HA) with NaOH will give the reaction equation:
[tex]\rm HA\;+\;NaOH\;\rightarrow\;NaA\;+H_2O[/tex]
This states that equal moles of NaOH and HA are required in the neutralization reaction.
For determining the concentration of the monoprotic acid, the HA is titrated against 0.2 M, 45 ml NaOH.
So, the concentration of the 25 ml HA, which is titrated will be;
[tex]\rm M_1\;V_1\;=\;M_2\;V_2[/tex]
[tex]\rm M_1\;and\;V_1[/tex] are molarity and volume of NaOH
[tex]\rm M_2\;and\;V_2[/tex] are molarity and volume of HA.
[tex]\rm 0.2\;\times\;45\;=\;M_2\;\times\;25[/tex]
[tex]\rm M_2[/tex] = 0.36 M
The concentration of monoprotic acid titrated against 0.2 M of 45 ml NaOH was found to be 0.36 M.
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